Question 12 Marks
Rationalise the denominator of the following: $\frac{1}{\sqrt{5}-2}$
AnswerIf a and b are integers, then $\big(\text{a}+\sqrt{\text{b}}\big)$ and $\big(\text{a}-\sqrt{\text{b}}\big)$ are rationalising factor of each other, as $\big(\text{a}+\sqrt{\text{b}}\big)\big(\text{a}-\sqrt{\text{b}}\big)=\big(\text{a}^2-\text{b}\big),$ which is rational.
Therefore, we have, $=\frac{1}{\big(\sqrt{5}-2\big)}=\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}$
$=\frac{\sqrt{5}+2}{\big(\sqrt{5}\big)^2-(2)^2}=\frac{\sqrt{5}+2}{5-4}$
$=\frac{\sqrt{5}+2}{1}=\sqrt{5}+2$
View full question & answer→Question 22 Marks
Examine whether the following number are rational or irrational:
$\sqrt{8}+4\sqrt{32}-6\sqrt{2}$
Answer $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$
$=\sqrt{4\times2}+4\sqrt{16\times2}-6\sqrt{2}$
$=2\sqrt{2}+16\sqrt{2}-6\sqrt{2}$
$=12\sqrt{2}$
Thus, the given number is irrational.
View full question & answer→Question 32 Marks
Examine whether the following numbers are rational or irrational.
$\sqrt{7}\times\sqrt{343}$
Answer As, $\sqrt{7}\times\sqrt{343}$
$=\sqrt{7\times343}$
$=\sqrt{2401}$
$=49,$ which is an integer
Hence, $\sqrt{7}\times\sqrt{343}$ is rational.
View full question & answer→Question 42 Marks
Multiply:
$18\sqrt{21}$ by $6\sqrt{7}$
Answer $18\sqrt{21}$ by $6\sqrt{7}$
$18\sqrt{21}\div6\sqrt{7}=\frac{18\sqrt{21}}{6\sqrt{7}}=\frac{3\sqrt{21}}{\sqrt{7}}=\frac{3\sqrt{21}\times\sqrt{7}}{\sqrt{7}\times\sqrt{7}}$
$=\frac{3\sqrt{3\times7\times7}}{7}=\frac{3\times7\sqrt{3}}{7}=3\sqrt{3}$
View full question & answer→Question 52 Marks
Simplify: $(1296)^\frac{1}{4}\times(1296)^\frac{1}{2}$
Answer$(1296)^\frac{1}{4}\times(1296)^\frac{1}{2}$ $=(6^4)^\frac{1}{4}\times(6^4)^\frac{1}{2}$ $=6^{4\times\frac{1}{4}}\times6^{4\times\frac{1}{2}}$ $=6\times6^2$ $=6\times36$ $=216$
View full question & answer→Question 62 Marks
Write the following in decimal form and say what kind of decimal expansion has. $\frac{15}{13}$
Answer$\frac{15}{13}=0.\overline{384615}$
It is a non-terminating recurring decimal. View full question & answer→Question 72 Marks
Multiply: $3\sqrt{5}$ by $2\sqrt{5}$
Answer$3\sqrt{5}$ by $2\sqrt{5}$ $3\sqrt{5}\times2\sqrt{5}=3\times2\times\sqrt{5}\times\sqrt{5}$ $=(3\times2\times5)=30$
View full question & answer→Question 82 Marks
Examine whether the following number are rational or irrational: $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$
Answer$\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$ $=\frac{2\sqrt{13}}{3\sqrt{4\times13}-4\sqrt{9\times13}}$ $=\frac{2\sqrt{13}}{3\times2\sqrt{13}-4\times3\sqrt{13}}$ $=\frac{2\sqrt{13}}{6\sqrt{13}-12\sqrt{13}}$ $=\frac{2\sqrt{13}}{-6\sqrt{13}}$ $=-\frac{1}{3}$ Thus, the given number is rational.
View full question & answer→Question 92 Marks
Give an example of a number x such that $x^2$ is an irrational number and $x^3$ is a rational number.
AnswerThe cube roots of natural numbers which are not perfect cubes are all irrational numbers. Let $\text{x}=\sqrt[3]{2}=2^{\frac{1}{3}}$ Now, $\text{x}^2=\big(2^{\frac{1}{3}}\big)^2=2^{\frac{2}{3}}=\big(2^2\big)^{\frac{1}{3}}=4^\frac{1}{3},$ which is an irrational number Also, $\text{x}^3=\Big(2^{\frac{1}{3}}\Big)^3=2^{3\times\frac{1}{3}}=2,$ which is a rational number.
View full question & answer→Question 102 Marks
Write the following in decimal form and say what kind of decimal expansion has. $\frac{3}{11}$
Answer$\frac{3}{11}=0.\overline{27}$
It is a non-terminating recurring decimal. View full question & answer→Question 112 Marks
Simplify the product $\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}.$
Answer$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}$ $=2^\frac{1}{3}\times2^\frac{1}{4}\times32^\frac{1}{12}$ $=2^\frac{1}{3}\times2^\frac{1}{4}\times2^{5\times\frac{1}{12}}$ $=2^\frac{1}{3}\times2^\frac{1}{4}\times2^\frac{5}{12}$ $=2^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$ $=2^{\frac{4+3+5}{12}}$ $=2^\frac{12}{12}$ $=2$
View full question & answer→Question 122 Marks
Examine whether the following numbers are rational or irrational. $3+\sqrt{3}$
AnswerLet us assume, to the contrary, that $3+\sqrt{3}$ is rational.
Then, $3+\sqrt{3}=\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are coprime and $\text{q}\neq0.$
$\Rightarrow\sqrt{3}=\frac{\text{p}}{\text{q}}-3$
$\Rightarrow\sqrt{3}=\frac{\text{p}-3\text{q}}{\text{q}}$ Since, $p$ and $q$ are are integers.
$\Rightarrow\frac{\text{p}-3\text{q}}{\text{q}}$ is rational.
So, $\sqrt{3}$ is also rational. But this contradicts the fact that $\sqrt{3}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $3+\sqrt{3}$ is rational.
Hence, $3+\sqrt{3}$ is irrational.
View full question & answer→Question 132 Marks
Find two rational numbers of the form $\frac{\text{p}}{\text{q}}$ between the numbers $0.2121121112...$ and$ 0.2020020002...$
AnswerThe rational numbers between the numbers $0.2121121112...$ and $0.2020020002...$ are: $0.21=\frac{21}{100}$ and $0.205=\frac{206}{1000}=\frac{41}{200}$
Disclaimer: There are an infinite number of rational numbers between two irrational numbers.
View full question & answer→Question 142 Marks
Simplify: $\Big(\frac{7776}{243}\Big)^{-\frac{3}{5}}$
Answer$\Big(\frac{7776}{243}\Big)^{-\frac{3}{5}}$ $=\Big(\frac{243}{7776}\Big)^{\frac{3}{5}}$ $=\Big(\frac{3^5}{6^5}\Big)^{\frac{3}{5}}$ $=\frac{3^{5\times\frac{3}{5}}}{6^{5\times\frac{3}{5}}}$ $=\frac{3^3}{6^3}$ $=\frac{3\times3\times3}{6\times6\times6}$ $=\frac{1}{8}$
View full question & answer→Question 152 Marks
Simplify: $\sqrt{72}+\sqrt{800}-\sqrt{18}$
Answer$\sqrt{72}+\sqrt{800}-\sqrt{18}$ $=\sqrt{ 36\times2}+\sqrt{400\times2}-\sqrt{9\times2}$ $=6\sqrt{2}+20\sqrt{2}-3\sqrt{2}$ $=23\sqrt{2}$
View full question & answer→Question 162 Marks
If $10^x = 64$, find the value of $10^{\big(\frac{\text{x}}{2}+1\big)}.$
AnswerGiven, $10^x$
= 64 $10^{\big(\frac{\text{x}}{2}+1\big)}$
$=10^\frac{\text{x}}{2}\times10^1$
$=(10\text{x})^\frac{1}{2}\times10$
$=(64)^\frac{1}{2}\times10$
$=(8^2)^\frac{1}{2}\times10$
$=8\times10$
$=80$
View full question & answer→Question 172 Marks
Find a rational number between $1.3$ and $1.4$
Answer$1.3$ and $1.4$
Let $x = 1.3$ and $y = 1.4$
Rational number lying between $x$ and $y$. $\frac{1}{2}(\text{x}+\text{y})=\frac{1}{2}\big(1.3+1.4\big)$
$=\frac{1}{2}(2.7)=1.35$
View full question & answer→Question 182 Marks
Simplify: $\big(3-\sqrt{3}\big)^2$
Answer $\big(3-\sqrt{3}\big)^2$ $=(3)^2+\big(\sqrt{3}\big)^2-2.3.\sqrt{3}$ $=9+3-6\sqrt{3}$ $=12-6\sqrt{3}$
View full question & answer→Question 192 Marks
Examine whether the following numbers are rational or irrational. $\sqrt{\frac{13}{117}}$
Answer$\sqrt{\frac{13}{117}}=\sqrt{\frac{1}{9}}=\frac{1}{3},$ which is rational Hence, $\sqrt{\frac{13}{117}}$ is rational.
View full question & answer→Question 202 Marks
Simplify: $\big(\sqrt{5}-\sqrt{3}\big)^2$
Answer$\big(\sqrt{5}-\sqrt{3}\big)^2$ $=\big(\sqrt{5}\big)^2+\big(\sqrt{3}\big)^2-2\sqrt{5}.\sqrt{3}$ $=5+3-2\sqrt{15}$ $=8-2\sqrt{15}$
View full question & answer→Question 212 Marks
Find the value of x in the following:
$\sqrt[5]{5\text{x}+2}=2$
Answer$\sqrt[5]{5\text{x}+2}=2$
$\Rightarrow(5\text{x}+2)^\frac{1}{5}=2$
$\Rightarrow\bigg[(5\text{x}+2)^\frac{1}{5}\bigg]=2^5$
$\Rightarrow5\text{x}+2=32$
$\Rightarrow5\text{x}=30$
$\Rightarrow\text{x}=6$
View full question & answer→Question 222 Marks
Rationalise the denominator of the following: $\frac{1}{\sqrt{7}-\sqrt{6}}$
Answer$\frac{1}{\sqrt{7}-\sqrt{6}}$ $=\frac{1}{\sqrt{7}-\sqrt{6}}\times\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$ $=\frac{\sqrt{7}+\sqrt{6}}{\big(\sqrt{7}\big)^2-\big(\sqrt{6}\big)^2}$ $=\frac{\sqrt{7}+\sqrt{6}}{7-6}$ $=\sqrt{7}+\sqrt{6}$
View full question & answer→Question 232 Marks
Write the following in decimal form and say what kind of decimal expansion has. $2\frac{5}{12}$
Answer$2\frac{5}{12}=\frac{29}{12}=2.41\overline{6}$ By actual division, we have:
It is a non-terminating decimal expansion. View full question & answer→Question 242 Marks
Prove that: $\frac{\text{x}^{\text{a}(\text{b}-\text{c})}}{\text{x}^{\text{b}(\text{a}-\text{c})}}\div\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{a}}\Big)^\text{c}=1$
Answer$\text{L.H.S}=\frac{\text{x}^{\text{a}(\text{b}-\text{c})}}{\text{x}^{\text{b}(\text{a}-\text{c})}}\div\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{a}}\Big)^\text{c}$ $=\frac{\text{X}^{\text{ab}-\text{ac}}}{\text{X}^{\text{ab}-\text{bc}}}\div\frac{\text{X}^\text{bc}}{\text{X}^\text{ac}}$ $=\text{X}^{\text{ab}-\text{ac}-\text{ab}+\text{bc}}\div\text{X}^{\text{bc}-\text{ac}}$ $=\text{X}^{\text{bc}-\text{ac}}\div\text{X}^{\text{bc}-\text{ac}}$ $=1$ $=\text{R.H.S}$
View full question & answer→Question 252 Marks
Simplify: $\Bigg(\frac{12^\frac{1}{5}}{27^\frac{1}{5}}\Bigg)^\frac{5}{2}$
Answer$\Bigg(\frac{12^\frac{1}{5}}{27^\frac{1}{5}}\Bigg)^\frac{5}{2}$$=\frac{12^{\frac{1}{5}\times\frac{5}{2}}}{15^{\frac{1}{5}\times\frac{5}{2}}}$
$=\frac{12^\frac{1}{2}}{27^\frac{1}{2}}$
$=\frac{\sqrt{12}}{\sqrt{27}}$
$=\frac{\sqrt{4\times3}}{\sqrt{9\times3}}$
$=\frac{2\sqrt3}{3\sqrt3}$
$=\frac{2}{3}$
View full question & answer→Question 262 Marks
Find the value of $x$ in the following:
$\frac{3^{3\text{x}}\times3^{2\text{x}}}{3^\text{x}}=\sqrt[4]{3^{20}}$
Answer$\frac{3^{3\text{x}}\times3^{2\text{x}}}{3^\text{x}}=\sqrt[4]{3^{20}}$
$\Rightarrow\frac{3^{3\text{x}+2\text{x}}}{3^\text{x}}=3^{20\times\frac{1}{4}}$
$\Rightarrow\frac{3^{5\text{x}}}{3^\text{x}}=3^5$
$\Rightarrow3^{4\text{x}}=3^5$
$\Rightarrow4\text{x}=5$
$\Rightarrow\text{x}=\frac{5}{4}$
View full question & answer→Question 272 Marks
Let a be a rational number and b be an irrational number. Is ab necessarily an irrational number? Justify your answer with an example.
Answera be a rational number and b be an irrational number then ab necessarily will be an irrational number. Example: $6$ is a rational number but $\sqrt{5}$ is irrational. And $6\sqrt{5}$ is also an irrational number.
View full question & answer→Question 282 Marks
Multiply: $\sqrt{10}$ by $\sqrt{40}$
Answer $\sqrt{10}$ by $\sqrt{40}$ $\sqrt{10}\times\sqrt{40}=\sqrt{10\times40}$ $=\sqrt{2\times5\times2\times2\times2\times5}$ $=(2\times2\times5)=20$
View full question & answer→Question 292 Marks
It being given that $\sqrt{3}=1.732,\sqrt{5}=2.236,\sqrt{6}=2.449$ and $\sqrt{10}=3.162,$ find to three places of decimal, the value of the following: $\frac{1}{\sqrt{6}+\sqrt{5}}$
Answer $\frac{1}{\sqrt{6}+\sqrt{5}}$$=\frac{1}{\sqrt{6}+\sqrt{5}}\times\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}$
$=\frac{\sqrt{6}-\sqrt{5}}{\big(\sqrt{6}\big)^2-\big(\sqrt{5}\big)^2}$
$=\frac{\sqrt{6}-\sqrt{5}}{6-5}$
$=\frac{\sqrt{6}-\sqrt{5}}{1}$
$=\sqrt{6}-\sqrt{5}$
$=2.449-2.236$
$=0.213$
View full question & answer→Question 302 Marks
If $a = 2, b = 3$, find the values of: $\big(\text{a}^{\text{a}}+\text{b}^{\text{b}}\big)^{-1}$
AnswerGiven, $a = 2$ and $b = 3$
$\big(\text{a}^{\text{a}}+\text{b}^{\text{b}}\big)^{-1}=(2^2+3^3)^{-1}$
$=(4+27)^{-1}$
$=(31)^{-1}$
$=\frac{1}{31}$
View full question & answer→Question 312 Marks
Simplify: $\Bigg(\frac{15^\frac{1}{3}}{9^\frac{1}{4}}\Bigg)^{-6}$
Answer$\Bigg(\frac{15^\frac{1}{3}}{9^\frac{1}{4}}\Bigg)^{-6}$ $=\Bigg(\frac{9^\frac{1}{4}}{15^\frac{1}{3}}\Bigg)^6$ $=\Bigg(\frac{3^{2\times\frac{1}{4}}}{15^\frac{1}{3}}\Bigg)^6$ $=\Bigg(\frac{3^\frac{1}{2}}{15^\frac{1}{3}}\Bigg)^6$ $=\frac{3^{\frac{1}{2}\times6}}{15^{\frac{1}{3}\times6}}$ $=\frac{3^3}{15^2}$ $=\frac{27}{225}$
View full question & answer→Question 322 Marks
Simplify: $\Bigg(\frac{15^\frac{1}{4}}{3^\frac{1}{2}}\Bigg)^{-2}$
Answer$\Bigg(\frac{15^\frac{1}{4}}{3^\frac{1}{2}}\Bigg)^{-2}$ $=\Bigg(\frac{3^\frac{1}{2}}{15^\frac{1}{4}}\Bigg)^2$ $=\frac{3^{\frac{1}{2}\times2}}{15^{\frac{1}{4}\times2}}$ $=\frac{3}{15^\frac{1}{2}}$
View full question & answer→Question 332 Marks
Find the value of x in the following: $\Big(\frac{3}{4}\Big)^3\Big(\frac{4}{3}\Big)^{-7}=\Big(\frac{3}{4}\Big)^{2\text{x}}$
Answer$\Big(\frac{3}{4}\Big)^3\Big(\frac{4}{3}\Big)^{-7}=\Big(\frac{3}{4}\Big)^{2\text{x}}$
$\Rightarrow\Big(\frac{3}{4}\Big)^3\Big(\frac{3}{4}\Big)^7=\Big(\frac{3}{4}\Big)^{2\text{x}}$
$\Rightarrow\Big(\frac{3}{4}\Big)^{3+7}=\Big(\frac{3}{4}\Big)^{2\text{x}}$
$\Rightarrow\Big(\frac{3}{4}\Big)^{10}=\Big(\frac{3}{4}\Big)^{2\text{x}}$
$\Rightarrow2\text{x}=10$
$\Rightarrow\text{x}=5$
View full question & answer→Question 342 Marks
Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}.$
AnswerAs, $\frac{5}{7}\approx0.714$ and $\frac{9}{11}\approx0.818$
So, the three different irrational numbers are: $0.72020020002..., 0.7515511555111...$ and $0.808008000...$
Disclaimer: There are an infinite number of irrational numbers between two rational numbers.
View full question & answer→Question 352 Marks
Simplify: $\Big(\frac{32}{243}\Big)^{-\frac{4}{5}}$
Answer$\Big(\frac{32}{243}\Big)^{-\frac{4}{5}}$ $=\Big(\frac{243}{32}\Big)^{\frac{4}{5}}$ $=\Big(\frac{3^5}{2^5}\Big)^{\frac{4}{5}}$ $=\frac{3^{5\times\frac{4}{5}}}{2^{5\times\frac{4}{5}}}$ $=\frac{3^4}{2^4}$ $=\frac{81}{16}$
View full question & answer→Question 362 Marks
Multiply: $2\sqrt{6}$ by $3\sqrt{3}$
Answer$2\sqrt{6}$ by $3\sqrt{3}$
$2\sqrt{6}\times3\sqrt{3}=2\times3\times\sqrt{6}\times\sqrt{3}$
$=6\times\sqrt{6\times3}$
$=6\times\sqrt{2\times3\times3}$
$=6\times3\sqrt{2}=18\sqrt{2}$
View full question & answer→Question 372 Marks
Write the following in decimal form and say what kind of decimal expansion has.
$\frac{231}{625}$
Answer $\frac{231}{625}=0.3696$ 
It is a terminating decimal expansion. View full question & answer→Question 382 Marks
Rationalise the denominator of the following: $\frac{4}{\sqrt{11}-\sqrt{7}}$
Answer$\frac{4}{\sqrt{11}-\sqrt{7}}$
$=\frac{4}{\sqrt{11}-\sqrt{7}}\times\frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}}$
$=\frac{4\big(\sqrt{11}+\sqrt{7}\big)}{\big(\sqrt{11}\big)^2-\big(\sqrt{7}\big)^2}$
$=\frac{4\big(\sqrt{11}+\sqrt{7}\big)}{11-7}$
$=\frac{4\big(\sqrt{11}+\sqrt{7}\big)}{4}$
$=\sqrt{11}+\sqrt{7}$
View full question & answer→Question 392 Marks
What are irrational numbers? How do they differ from rational numbers? Give examples.
AnswerA number that can neither be expressed as a terminating decimal nor be expressed as a repeating decimal is called an irrational number. A rational number, on the other hand, is always a terminating decimal, and if not, it is a repeating decimal.
Examples of irrational numbers:
$0.101001000...$
$0.232332333...$
View full question & answer→Question 402 Marks
On her birthday Reema distributed chocolates in an orphanage. The total number of chocolates she distributed is given by $\big(5+\sqrt{11}\big)\big(5-\sqrt{11}\big).$
$i.$ Find the number of chocolates distributed by her.
$ii. $Write the moral values depicted here by Reema.
Answer$i.$ Number of chocolates distributed by Reema
$=\big(5+\sqrt{11}\big)\big(5-\sqrt{11}\big)$
$=(5)^2-\big(\sqrt{11}\big)^2$
$=25-11$
$=14$
$ii.$ Loving, helping and caring attitude towards poor and needy children.
View full question & answer→Question 412 Marks
Find a rational number between -1 and $\frac{1}{2}$
Answer$-1$ and $\frac{1}{2}$
Let:
$\text{x}=-1$ and $\text{y}=\frac{1}{2}$
Rational number lying between $x$ and $y.$
$\frac{1}{2}(\text{x}+\text{y})=\frac{1}{2}\Big(-1+\frac{1}{2}\Big)$
$=-\frac{1}{4}$
View full question & answer→Question 422 Marks
Write the following in decimal form and say what kind of decimal expansion has. $\frac{5}{8}$
Answer$\frac{5}{8}=0.625$ By actual division, we have: 
It is a terminating decimal expansion. View full question & answer→Question 432 Marks
Multiply: $3\sqrt{28}$ by $2\sqrt{7}$
Answer$3\sqrt{28}$ by $2\sqrt{7}$
$3\sqrt{28}\times2\sqrt{7}=3\times2\times\sqrt{28}\times\sqrt{7}$
$=6\times\sqrt{28\times7}$
$=6\times\sqrt{2\times2\times7\times7}$
$=(6\times2\times4)=84$
View full question & answer→Question 442 Marks
Examine whether the following number are rational or irrational:
$\big(\sqrt{3}+2\big)^2$
Answer $\big(\sqrt{3}+2\big)^2$
$=\big(\sqrt{3}\big)^2+2\times2\times\sqrt{3}+(2)^2$
$=3+4\sqrt{3}+4$
$=7+4\sqrt{3}$
Clearly, the given number is irrational.
View full question & answer→Question 452 Marks
How many irrational numbers lie between $\sqrt{2}$ and $\sqrt{3}?$ Find any three irrational numbers lying between $\sqrt{2}$ and $\sqrt{3}.$
AnswerThere are infinite number of irrational numbers lying between $\sqrt{2}$ and $\sqrt{3}.$
As, $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$
So, the three irrational numbers lying between $\sqrt{2}$ and $\sqrt{3}$ are: $1.420420042000...,$ $1.505005000...$ and $1.616116111...$
View full question & answer→Question 462 Marks
Write the following in decimal form and say what kind of decimal expansion has. $\frac{11}{24}$
Answer$\frac{11}{24}=0.458\overline{3}$ By actual division, we have:
It is a non-terminating recurring decimal expansion. View full question & answer→Question 472 Marks
Is the product of two irrationals always irrational? Justify your answer.
AnswerProduct of two irrational numbers is not always an irrational number. Example: $\sqrt{5}$ is irrational number. And $\sqrt{5}\times\sqrt{5}=5$ is a rational number. But the product of another two irrational numbers $\sqrt{2}$ and $\sqrt{3}$ is $\sqrt{6}$ which is also an irrational numbers.
View full question & answer→Question 482 Marks
Simplify $\Bigg[\Big\{(256)^{-\frac{1}{2}}\Big\}^{-\frac{1}{4}}\Bigg]^2.$
Answer$\Bigg[\Big\{(256)^{-\frac{1}{2}}\Big\}^{-\frac{1}{4}}\Bigg]^2$
$=\Bigg[\Big\{\big(16^2\big)^{-\frac{1}{2}}\Big\}^{-\frac{1}{4}}\Bigg]^2$
$=\Bigg[\Big\{16^{-1}\Big\}^{-\frac{1}{4}}\Bigg]^2$
$=\Bigg[16^{-1\times\big(-\frac{1}{4}\big)}\Bigg]^2$
$=\Bigg[16^{\frac{1}{4}}\Bigg]^2$
$=\Bigg[2^{4\times\frac{1}{4}}\Bigg]^2$
$=2^2$
$=4$
View full question & answer→Question 492 Marks
Find two rational and two irrational number between $0.5$ and $0.55.$
AnswerThe two rational numbers between $0.5$ and $0.55$ are: $0.51$ and $0.52$
The two irrational numbers between $0.5$ and $0.55$ are: $0.505005000...$ and $0.5101100111000...$
Disclaimer: There are infinite number of rational and irrational numbers between $0.5$ and $0.55.$
View full question & answer→Question 502 Marks
Multiply: $6\sqrt{15}$ by $4\sqrt{3}$
Answer$6\sqrt{15}$ by $4\sqrt{3}$
$6\sqrt{15}\times4\sqrt{3}=6\times4\times\sqrt{15}\times\sqrt{3}$
$=24\times\sqrt{15\times3}$
$=24\times\sqrt{3\times5\times3}$
$=24\times3\sqrt{5}=72\sqrt{5}$
View full question & answer→